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If a cos θ + b sin θ = m and a sin θ - b cosθ = n,
then show that a² + b²-m² + n²

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Answer 1

Answer:

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Step-by-step explanation:

On squaring Eqs. (i) and (ii) and then adding the resulting equations, we get

m² + n² = (a cos 0+ b sin 0)² + (a sin 8-b cos 0)²

=a² cos² e + b² sin² 0 + 2ab sin e-cos + a² sin² 0

+ b² cos² 0-2ab sin 8- cos 0

=a²(cos² 0+ sin² ) + b² (sin² 0 + cos² e) = a² + b²

Answer 2

♡αղsաҽɾ:−)

Given,

a cos θ + b sin θ = m …(1)

a sin θ – b cos θ = n …(2)

Squaring and adding equation 1 and 2, we get

(a cos θ + b sin θ)2 + (a sin θ – b cos θ)2

= m2 + n2

⇒ a2cos2θ + b2sin2θ + 2ab sin θ cos θ + a2sin2θ + b2cos2θ - 2ab sin θ cos θ

= m2 + n2

⇒ a2cos2θ + b2sin2θ + a2sin2θ + b2cos2θ

= m2 + n2

⇒ a2(sin2θ + cos2θ) + b2(sin2θ + cos2θ)

= m2 + n2 Using: sin2θ + cos2θ = 1, we get –

⇒ a2 + b2

=m2+ n2

Using: sin2θ + cos2θ = 1, we get

= a2 + b2 = m2 + n2

【Hope it helps uhh】